Last week…

• Correlation r: a measure of the strength and direction of the linear relationship between two variables
• Is there a causal relationship between two variables?
  • No: use correlation analysis
  • Yes: use regression analysis

Fitting the line

Y_i = \beta_0 + \beta_1 x_i + \epsilon_i

Y = c + mx + \epsilon

How do we fit a line to data if data are “noisy”?

x <- 1:10
y <- 2 * x + rnorm(10, 0, 2)
# generate y with predicted values
y_pred <- 2 * x
df <- data.frame(x, y)

p1 <- ggplot(df, aes(x, y_pred)) +
  geom_point(size = 2) +
  geom_smooth(method = "lm", se = FALSE, color = "firebrick") +
  labs(x = "x", y = "y", title = "A")

p2 <- ggplot(df, aes(x, y)) +
  geom_point(size = 2) +
  geom_smooth(method = "lm", se = FALSE, color = "royalblue") +
  labs(x = "x", y = "y", title = "B (How do we fit this?)")

library(patchwork)
p1 + p2 + plot_layout(ncol = 2)

Usage

• Student’s t-test
• linear regression

Galton’s data revisited

• Galton’s data on the heights of parents and their children.
• Is there a relationship between the heights of parents and their children?

library(HistData)
data(Galton)
fit <- lm(child ~ parent, data = Galton)
ggplot(Galton, aes(x = parent, y = child)) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE, color = "firebrick") +
  labs(x = "Parent height (inches)", y = "Child height (inches)")

How did we end up with the line in the plot above?

How do we fit a line?

• Minimise the sum of the squared residuals:

\color{firebrick}{argmin_{\beta_0, \beta_1}} \sum_{i=1}^n (y_i - \color{royalblue}{(\beta_0 + \beta_1 x_i)})^2

Source

Residuals, \hat \epsilon

\color{firebrick}{\hat{\epsilon_i}} = \color{royalblue}{y_i} - \color{forestgreen}{\hat{y_i}}

# simulate example data
set.seed(340)
x <- runif(8, 0, 30)
y <- 5 * x + rnorm(8, 0, 40)
df <- data.frame(x, y)

# fit linear model, add residual vertical lines as arrows
mod <- lm(y ~ x, data = df)
p1 <- ggplot(df, aes(x, y)) +
  geom_point() +
  geom_segment(aes(xend = x, yend = fitted(mod)),
    arrow = arrow(length = unit(0.2, "cm")),
    color = "royalblue"
  ) +
  labs(x = "x", y = "y")

p1 +
  geom_smooth(method = "lm", se = FALSE, color = "firebrick") +
  annotate("text",
    x = 6.3, y = -6, size = 7,
    label = expression(hat(epsilon[i])), colour = "royalblue"
  ) +
  annotate("text",
    x = 5.6, y = 25, size = 7,
    label = expression(hat(y[i])), colour = "forestgreen"
  ) +
  annotate("text",
    x = 5.6, y = -36, size = 7,
    label = expression(y[i]), colour = "firebrick"
  )

Slope, \beta_1

\beta_1 = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2} = \frac{Cov(x,y)}{Var(x)} = \frac{SS_{xy}}{SS_{xx}}

# Calculate slope from df
beta1 <- sum((df$x - mean(df$x)) * (df$y - mean(df$y))) /
  sum((df$x - mean(df$x))^2)
# beta0 <- mean(df$y) - beta1 * mean(df$x)

p1 +
  geom_smooth(method = "lm", se = FALSE, color = "firebrick", linetype = 2) +
  # label the line
  annotate("text",
    x = 15, y = 65, size = 7,
    label = expression(beta[1]), colour = "firebrick"
  )

Intercept

\beta_0 = \bar{y} - \beta_1 \bar{x}

# calculate mean y from df
ybar <- mean(df$y)
xbar <- mean(df$x)
beta0 <- ybar - beta1 * xbar

p1 + geom_vline(xintercept = xbar, linetype = "dashed", color = "slateblue") +
  geom_hline(yintercept = ybar, linetype = "dashed", color = "slateblue") +
  # label the lines
  annotate("text",
    x = 25, y = ybar * 0.8, size = 7,
    label = expression(bar(y)), colour = "slateblue"
  ) +
  annotate("text",
    x = xbar * 1.05, y = 150, size = 7,
    label = expression(bar(x)), colour = "slateblue"
  ) +
  geom_smooth(method = "lm", se = FALSE, color = "firebrick", linetype = 2) +
  # extend the geom_smooth line to intercept x=0
  geom_segment(aes(x = xbar, y = ybar, xend = 0, yend = beta0),
    color = "firebrick", linetype = 2
  ) +
  # label the slope line
  annotate("text",
    x = 15, y = 65, size = 7,
    label = expression(beta[1]), colour = "firebrick"
  ) +
  # add a dot at the intercept
  geom_point(aes(x = 0, y = beta0), color = "seagreen", size = 3) +
  # label the intercept
  annotate("text",
    x = 0, y = beta0 * 1.4, size = 7,
    label = expression(beta[0]), colour = "seagreen"
  )

Fitting a linear model in R

Is there a relationship between the heights of parents and their children?

fit <- lm(child ~ parent, data = Galton)
fit

Call:
lm(formula = child ~ parent, data = Galton)

Coefficients:
(Intercept)       parent  
    23.9415       0.6463  

\widehat{child} = 23.9 + 0.646 \cdot parent

Assumptions

The data must meet certain criteria, which we often call assumptions. They can be remembered using LINE:

• Linearity. The relationship between y and x is linear.
• Independence. The errors \epsilon are independent.
• Normal. The errors \epsilon are normally distributed.
• Equal Variance. At each value of x, the variance of y is the same i.e. homoskedasticity, or constant variance.

Assumptions: Why do we care?

• If the assumptions are met, then we can be confident that the model is a good representation of the data.
• If they are not met, the results are still presented, but our interpretation of the model is likely to be flawed.

How do we check the assumptions?

Recall that the linear model is a deterministic straight line equation y = c + mx plus some random noise \epsilon:

Y_i = \beta_0 + \beta_1 x + \epsilon

• If the only source of variation in y is \epsilon, then we can check our assumptions by just looking at the residuals \hat{\epsilon}.

How do we get the residuals?

• Fit the model…
• Residuals need to be calculated from the model, not from the raw data.
• In R, these values are stored automatically.

Another way to look at residuals

Once you have fitted the line, it does not change. The residuals are the vertical distances between the points (not shown) and the line.

1-step

par(mfrow = c(2, 2)) # need to do this to get 4 plots on one page
plot(fit)

• Residuals vs. Fitted: check for linearity, equal variance.
• Q-Q Residuals: check for normality.
• Scale-Location: check for equal variance (standardised).
• Residuals vs. Leverage: check for outliers (influential points).

Assumption: Linearity

• Residuals vs. fitted plot looks at the relationship between the residuals and the fitted values.
• If the relationship is linear:
  • Residuals should be randomly scattered around the horizontal axis.
  • The red line should be reasonably straight.

plot(fit, which = 1)

Assumption: Normality

• Q-Q plot looks at the distribution of the residuals against a normal distribution function (the dotted line).
• Sometimes, a histogram is still useful to see the shape of the distribution.

par(mfrow = c(1, 2))
plot(fit, which = 2)
hist(rstandard(fit))

Assumption: Normality

• If normally distributed, the points should follow the red line.
• Deviation from the red line is common in the tails (i.e. the ends), but not in the middle.

Equal variances

• Look at the scale-location plot.
• If variances are equal, the points should be randomly scattered around the horizontal axis.
• The red line should be more or less horizontal.

plot(fit, which = 3)

Equal variances

• If variances are not equal we may see:
  • A funnel shape, where the points are more spread out at the ends than in the middle. Sometimes also called “fanning”.
  • Patterns in the scale-location plot, such as a curve or a wave, indicating that the variance is changing.
• Look at the red line for a general trend, but don’t depend on it too much.

set.seed(915)
x <- rnorm(100)
y <- 2 + 5 * x^2 + rchisq(100, df = 2)
df <- data.frame(x, y)
fit_eg <- lm(y ~ x, data = df)
plot(fit_eg, which = 3)

Outliers

• Leverage is a measure of how far away the predictor variable is from the mean of the predictor variable.
• The Residuals vs Leverage plot shows the relationship between the residuals and the leverage of each point.
• Cook’s distance is a measure of how much the model would change if a point was removed.

In general, points with high leverage and high Cook’s distance are considered outliers.

plot(fit, which = 5)

Example of an influential outlier

set.seed(1028)
x <- rnorm(100)
y <- 2 + 5 * x + rnbinom(100, 10, .5)
y[60] <- y[60] + 30
df <- data.frame(x, y)
# Add an outlier
fit_eg <- lm(y ~ x, data = df)
plot(fit_eg, which = 5)

We don’t want points to exceed the dashed line (which appears once they approach the Cook’s distance), because that means they are likely to influence the model greatly.

It depends…

• Depends which assumption is not met and the type of data i.e. circumstances.
  • If data is non-linear, try a transformation of the response variable y, from light to extreme:
    • root: \sqrt{y} or \sqrt{y+1} if y contains zeros
    • log: \log(y) or \log(y+1) if y contains zeros
    • inverse: \frac{1}{y} or \frac{1}{y+1} if y contains zeros

What if that doesn’t work?

• If the assumptions are still not met after trying the above, you can try:
  • Using a different model e.g. generalized linear model.
  • Using a different type of regression e.g. logistic regression.
  • Using a non-parametric test.

Model assumptions are met. Now what?

summary(fit)

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

The model so far

library(HistData)
data(Galton)
fit <- lm(child ~ parent, data = Galton)
summary(fit)

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

Hypothesis testing

How does our null (H_0: \beta_1=0) model compare to the linear (H_0: \beta_1 \neq 0) model?

null_model <- Galton %>%
  lm(child ~ 1, data = .) %>%
  augment(Galton)
lin_model <- Galton %>%
  lm(child ~ parent, data = .) %>%
  augment(Galton)
models <- bind_rows(null_model, lin_model) %>%
  mutate(model = rep(c("Null model", "SLR model"), each = nrow(Galton)))

ggplot(data = models, aes(x = parent, y = child)) +
  geom_smooth(
    data = filter(models, model == "Null model"),
    method = "lm", se = FALSE, formula = y ~ 1, size = 0.5
  ) +
  geom_smooth(
    data = filter(models, model == "SLR model"),
    method = "lm", se = FALSE, formula = y ~ x, size = 0.5
  ) +
  geom_segment(
    aes(xend = parent, yend = .fitted),
    arrow = arrow(length = unit(0.1, "cm")),
    size = 0.3, color = "darkgray"
  ) +
  geom_point(alpha = .2) +
  facet_wrap(~model) +
  xlab("Parent height (in)") +
  ylab("Child height (in)")

What are we testing?

• The null model is a model with no predictors, i.e. y = \beta_0 + \epsilon
• The linear model is a model with one predictor, i.e. y = \beta_0 + \beta_1 x + \epsilon
• We use the t-test to compare the two models:

t = \frac{estimate - 0}{Standard\ error} = \frac{\hat{\beta}_1 - 0}{SE(\hat{\beta}_1)} where SE(\hat{\beta}_1) is the standard error of the slope estimate:

SE(\hat{\beta}_1) = \frac{\hat{\sigma}}{\sqrt{\sum_{i=1}^n (x_i - \bar{x})^2}}

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Call: the model formula

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Residuals: distribution of the residuals

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Coefficients: a summary table of the coefficients, their standard errors, t-values, and p-values addressing the hypothesis that the coefficient is 0

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• (Intercept) term is the mean of the response when all predictors are 0, which is not meaningful in most cases. In this case, it is the mean child height when the parent height is 0.
• parent: the slope coefficient that we are interested in, which represents the change in the mean of the response for a one-unit increase in the predictor.
  • The p-value (Pr) tells us whether the slope is significantly different from 0.
  • If it is, then we can conclude that there is a significant linear relationship between the predictor and the response.

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• We can also use the Estimate values to write the equation of the regression line: \widehat{child} = 23.94153 + 0.64629 \cdot parent

• For every one-inch increase in the parent height, the child height is predicted to increase by 0.64629 inches.

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Residual standard error: the standard deviation of the residuals.
  • Interpretation: the average amount that the response will deviate from the true regression line.
• degrees of freedom: the number of observations minus the number of parameters being estimated. Used in hypothesis testing and calculating the standard error of the regression coefficients.
  • Can estimate sample size from this number.

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Multiple R-squared: the proportion of variance explained by the model.
• Adjusted R-squared: the proportion of variance explained by the model, adjusted for the number of predictors.
• Interpretation:
  • Ranges from 0 to 1.
  • Since this is SLR, we can interpret this as the proportion of variance in the response that is explained by parent: 21.05% (from Multiple R-squared).

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• F-statistic: the ratio of the variance of the regression model to the variance of the residuals.
  • Also known as the partial F-test between the full model and the intercept-only (null) model.
• p-value: for the linear model, the p-value is the probability that the F-statistic is greater than the observed value under the null hypothesis.
  • A significant p-value indicates that the linear model is a better fit than the intercept-only model.

Two methods

fit <- lm(formula = child ~ parent, data = Galton)
anova(fit)

Analysis of Variance Table

Response: child
           Df Sum Sq Mean Sq F value    Pr(>F)    
parent      1 1236.9 1236.93  246.84 < 2.2e-16 ***
Residuals 926 4640.3    5.01                      
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

summary(fit)

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

Two methods

Explore

Read the data:

library(readxl) # load the readxl package

alligator <- read_excel(path = "data/ENVX1002_Lecture_wk10_data.xlsx", 
  sheet = "Alligator") # read in the data

What does the data look like?

str(alligator)

tibble [25 × 2] (S3: tbl_df/tbl/data.frame)
 $ Length: num [1:25] 58 61 63 68 69 72 72 74 74 76 ...
 $ Weight: num [1:25] 28 44 33 39 36 38 61 54 51 42 ...

Plot

plot(x = alligator$Length, y = alligator$Weight, 
  xlab = "Length (cm)", ylab = "Weight (kg)")

library(ggplot2) # load the ggplot2 package
ggplot(data = alligator, aes(x = Length, y = Weight)) +
  geom_point() +
  labs(x = "Length (cm)", y = "Weight (kg)")

Plot residual diagnostics

To check assumptions, we need to fit the model first, then plot the model.

fit <- lm(formula = Weight ~ Length, data = alligator)
par(mfrow = c(2, 2)) # set up a 2 x 2 grid for plots
plot(fit)

Check assumptions

Is the relationship linear?

plot(fit, which = 1)

If the linearity assumption is not met, there is no reason to validate the model since it is no longer suitable for the data.

Dealing with non-linearity: transform the data

library(patchwork)

p1 <- ggplot(data = alligator, aes(x = Length, y = Weight)) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "Weight (kg)", title = "Original") +
  geom_smooth(se = FALSE, linetype = 2)

p2 <- ggplot(data = alligator, aes(x = Length, y = sqrt(Weight))) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "sqrt[Weight (kg)]", title = "Square root") +
  geom_smooth(se = FALSE, linetype = 2)

p3 <- ggplot(data = alligator, aes(x = Length, y = log(Weight))) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "log[Weight (kg)]", title = "Natural log") +
  geom_smooth(se = FALSE, linetype = 2)

p4 <- ggplot(data = alligator, aes(x = Length, y = log10(Weight))) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "log10[Weight (kg)]", title = "Log base 10") +
  geom_smooth(se = FALSE, linetype = 2)

p1 + p2 + p3 + p4

Natural log transformation

fit <- lm(formula = log(Weight) ~ Length, data = alligator)
plot(fit, which = 1)

Natural log transformation – Check assumptions again

par(mfrow = c(2, 2)) # set up a 2 x 2 grid for plots
plot(fit)

Interpretation

summary(fit)

Call:
lm(formula = log(Weight) ~ Length, data = alligator)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.289266 -0.079989  0.000933  0.102216  0.288491 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 1.335335   0.131394   10.16 5.63e-10 ***
Length      0.035416   0.001506   23.52  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1493 on 23 degrees of freedom
Multiple R-squared:  0.9601,    Adjusted R-squared:  0.9583 
F-statistic:   553 on 1 and 23 DF,  p-value: < 2.2e-16

• Length is a statistically significant predictor of log(Weight) (p < .001).
• The model explains a statistically significant and large proportion (96%) of variance (R² = 0.96, F(1, 23) = 553, p < .001)
• For every 1 cm increase in Length, log(Weight) increases by 0.0354.
  • Or, for every 1 cm increase in Length, percent increase in Weight is 3.54% (only works when transforming using natural log).

Summary

You should know the workflow by now

1. Explore
2. Plot
3. Fit model and plot residual diagnostics
4. Check assumptions, transform data if necessary. Go back to step 3.
5. Interpret